Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 45062 Accepted Submission(s): 16339

Problem Description

Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).

Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).

But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^

Input

Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn. Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 3
2 6 -1 4 -2 3 -2 3

Sample Output

6
8

Hint
Huge input, scanf and dynamic programming is recommended.

Author

JGShining（极光炫影）

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Accepted Code

package cn.edu.hdu.acm;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

//最开始没读懂题意，后来网上搜了才知道是n个数中找最大的不相交的m子段和，dp
//转移方程： cur[i,j]=max(cur[i,j-1]+a[j],cur[i-1,t]+a[j]) 表示前j项中最大的i个字段的最大和


  /*设状态为 cur[i,j]，表示前 j 项分为 i 段的最大和，且第 i 段必须包含 data[j]，则状态转移方程如下：
cur[i,j] = max{cur[i,j − 1] + data[j],max{cur[i − 1,t] + data[j]}}, 其中i ≤ j ≤ n,i − 1 ≤ t < j
target = max{cur[m,j]}, 其中m ≤ j ≤ n
分为两种情况：
• 情况一，data[j] 包含在第 i 段之中，cur[i,j − 1] + data[j]。
• 情况二，data[j] 独立划分成为一段，max{cur[i − 1,t] + data[j]}。
观察上述两种情况可知 cur[i,j] 的值只和 cur[i,j-1] 和 cur[i-1,t] 这两个值相关，因此不需要二维数组，
可以用滚动数组，只需要两个一维数组，用 cur[j] 表示现阶段的最大值，即 cur[i,j − 1] + data[j]，用
pre[j] 表示上一阶段的最大值，即 max{cur[i − 1,t] + data[j]}。*/

public class Main1024 {
    static final int maxn=(int)1e6+5;
    static int a[]=new int[maxn];

    static int max(int a,int b){
        return a>b?a:b;
    }
    public static void main(String[] args) throws IOException {
        StreamTokenizer st=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        int m,n;
        while(st.nextToken()!=StreamTokenizer.TT_EOF){
            m=(int)st.nval;
            st.nextToken();
            n=(int)st.nval;
            for(int i=1;i<=n;i++){
                st.nextToken();
                a[i]=(int)st.nval;
            }
            int cur[]=new int[n+1];//表示现阶段的最大值
            int pre[]=new int[n+1];//表示上一个阶段的最大值
            int j=0;
            int max_sum=0;
            for(int i=1;i<=m;i++){
                max_sum=-0x3f3f3f3f;
                for(j=i;j<=n;j++){
                    cur[j]=max(cur[j-1]+a[j],pre[j-1]+a[j]);
                    pre[j-1]=max_sum;//保存上一个阶段的最大值
                    if(max_sum<cur[j]){
                        max_sum=cur[j];//记录这个阶段的最大值
                    }
                }
                pre[j-1]=max_sum;//保存上一个阶段的最大值，因为j++了，所以这里是j-1
            }
            System.out.println(max_sum);
        }
    }
}